Examples:
$$
Non-examples:
$x^2+3 x$ (not written as a fraction) $\sqrt{x}+1$ (not a ratio of polynomials) We treat these like numerical fractions, but we must also pay attention to values that make the denominator zero .
A rational expression is undefined when its denominator is zero.
$$
The denominator is zero when $x-2=0 \Rightarrow x=2$. So the expression is defined for all real $x$ except $x=2$. We can write the restriction as: $$
$$
Denominator is zero when $x=0$ or $x-1=0 \Rightarrow x=1$. Restrictions: $x \neq 0,1$. Always state restrictions before simplifying or cancelling any factors, because simplification can hide the values that make the original denominator zero.
To simplify a rational expression:
Factor the numerator and denominator (if possible).Cancel any common factors (not terms!) that appear in both.State the restrictions on the variable Simplify $\frac{x^2-9}{x^2+4 x+3}$.
Solution
Factor: $$x^2-9=(x-3)(x+3), \quad x^2+4 x+3=(x+1)(x+3) $$ Cancel the common factor $x+3$ : $$\frac{(x-3)(x+3)}{(x+1)(x+3)}=\frac{x-3}{x+1} $$ Restrictions from the original denominator:$$x+1 \neq 0 \Rightarrow x \neq-1 \text { and } x+3 \neq 0 \Rightarrow x \neq-3$$ Final answer:$$\frac{x^2-9}{x^2+4 x+3}=\frac{x-3}{x+1}, \quad x \neq-1,-3$$ Cancelling terms instead of factors , e.g. trying to cancel the $x^2$ in $\frac{x^2+4}{x^2}$. You may only cancel whole factors multiplied, not added. These are the easiest operations: treat them like numerical fractions.
$$\frac{a}{b} \times \frac{c}{d}=\frac{a c}{b d}, \quad b \neq 0, d \neq 0$$
Steps :
Factor all numerators and denominators. Cancel any common factors. Multiply what remains. State restrictions from all original denominators. $$\frac{x}{x-2} \times \frac{x-2}{x+3}$$
Factor: already factored. Cancel the common factor $x-2$ : $$\frac{x}{x-2} \times \frac{x-2}{x+3}=\frac{x}{x+3}, \quad x \neq 2,-3 $$ We must still exclude $x=2$ even though $x-2$ cancelled. $$\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \times \frac{d}{c}, \quad b \neq 0, c \neq 0, d \neq 0$$
So we multiply by the reciprocal of the second fraction.
$$\frac{2 x}{x-3} \div \frac{x+1}{3 x}$$
Rewrite as multiplication: $$\frac{2 x}{x-3} \times \frac{3 x}{x+1} .$$ Factor (if needed) and cancel:$$\frac{2 x}{x-3} \times \frac{3 x}{x+1}=\frac{6 x^2}{(x-3)(x+1)} .$$ Restrictions: $x \neq 0,3,-1$ (from all denominators, including the original divisor). For division questions, never divide across ; always “multiply by the reciprocal” first, then simplify.
To add or subtract fractions, you need a common denominator .
General idea:
$$\frac{A}{B}+\frac{C}{D}=\frac{A D+C B}{B D}, \quad B, D \neq$$
In practice, this means:
Factor denominators. Find the lowest common denominator (LCD) . Rewrite each fraction with the LCD. Combine numerators. Simplify if possible and state restrictions. $$\frac{3 x+1}{x-4}-\frac{x-5}{x-4}$$
Same denominator, so subtract numerators:
$$\frac{(3 x+1)-(x-5)}{x-4}=\frac{3 x+1-x+5}{x-4}=\frac{2 x+6}{x-4}=\frac{2(x+3)}{x-4}, \quad x \neq 4$$
$$\frac{1}{x}+\frac{2}{x+1}$$
LCD is $x(x+1)$. Rewrite:$$\frac{1}{x}=\frac{x+1}{x(x+1)}, \quad \frac{2}{x+1}=\frac{2 x}{x(x+1)} $$ Add:$$\frac{x+1}{x(x+1)}+\frac{2 x}{x(x+1)}=\frac{3 x+1}{x(x+1)} $$ Restrictions: $x \neq 0,-1$. Don't add denominators directly: $\frac{1}{x}+\frac{2}{x+1} \neq \frac{3}{2 x+1}$. You must use a common denominator . You may need to solve a formula for a different variable when that variable appears in a denominator.
General strategy :
Identify all terms containing the target variable. Clear denominators by multiplying both sides by the common denominator. Collect like terms and solve as usual. Make $x$ the subject of $\frac{a}{x}=b $.
Solution
Multiply both sides by $x$ : $$a=b x $$ Now solve for $x$ : $$x=\frac{a}{b}, \quad b \neq 0, x \neq 0 \text { in the original formula } $$ Make $x$ the subject of $y=\frac{3 x-2}{x+1}$.
Solution
Multiply both sides by $x+1$ : $$y(x+1)=3 x-2 $$ Expand the left: $$ y x+y=3 x-2 $$ Collect $x$-terms on one side: $$y x-3 x=-2-y $$ Factor out $x$ : $$x(y-3)=-(y+2) $$ Solve for $x$ : $$x=\frac{-(y+2)}{y-3}, \quad y \neq 3, x \neq-1 \text { in the original }(\text { since } x+1 \neq 0)$$ Graphing technology (on a calculator or graphing software) is useful for:
visualising where an expression is undefined (vertical gaps/asymptotes), checking if two algebraic expressions are equivalent for all allowed values, exploring how the graph changes when you cancel factors . In the next articles, you will learn about rational functions and equations, and how graphing technology helps with those.
State the restrictions and simplify:$$\frac{x^2-4}{x^2-3 x} $$ Simplify and give the restrictions:$$\frac{3}{x}+\frac{2}{x-1}$$ Make $x$ the subject:$$y=\frac{2 x+5}{x-3} $$ Explain why $\frac{x}{x+1} \neq 1$ for all $x$, even though the numerator and denominator both "contain $x$ ". Describe how the graph of $\frac{x^2-9}{x-3}$ is related to the graph of $y=x+3$, and explain any restrictions.